In the proof presented in my textbook, it utilises the equation

$$(A-\lambda I) \operatorname{Adj}(A - \lambda I) = \det(A-\lambda I) \, I$$

And it stated that,

$$P(A-\lambda I) = \det(A-\lambda I)= a_0 + a_1\lambda + ... + a_n\lambda^n$$

and that

$$Q(A-\lambda I) = \operatorname{Adj}((A-\lambda I) = b_0 + b_1\lambda + ... + b_k\lambda^k$$

I have no problem understanding that $P(A-\lambda I)$ is the characteristic polynomial for matrix $A$. However, I have problem understanding 2 things,

  1. Why $\operatorname{Adj}(A-\lambda I)$ can be written into a polynomial?

  2. What does it mean to then multiply the above polynomial to matrix $(A-\lambda I)$?

Many thanks!

  • $\begingroup$ You can answer your questions for $n=2$ yourself as follows. Write out all expressions explicitly for $A=\begin{pmatrix} a & b \cr c & d \end{pmatrix}$. How do you define $Adj$? You can compute it explicitly for $A$, see here. $\endgroup$ – Dietrich Burde Mar 2 at 9:25
  • $\begingroup$ I mean if I just do the operation $\begin{pmatrix} d-\lambda & -b \cr -c & a-\lambda \end{pmatrix}$ $\begin{pmatrix} a-\lambda & b \cr c & d-\lambda \end{pmatrix}$. What I get is $\begin{pmatrix} (a-\lambda)(d- \lambda) - bc & 0 \cr 0 & (a-\lambda)(d-\lambda) + bc \end{pmatrix}$. Correct? $\endgroup$ – Klein_Bottle Mar 2 at 11:08
  • 2
    $\begingroup$ Adj$(A-\lambda I)$ can be thought of as a polynomial in $\lambda$ whose coefficients are matrices. I think Adj$(A-\lambda I)=B_0+B_1\lambda+\cdots +B_{n-1}\lambda^{n-1}$ is a better way of writing it. $\endgroup$ – Lord Shark the Unknown Mar 2 at 11:11
  • $\begingroup$ The equality in your question should be $\color{red}{(A-\lambda I)}\operatorname{Adj}(A-\lambda I)=\det(A-\lambda I)I$, not $\color{red}{A}\operatorname{Adj}(A-\lambda I)=\det(A-\lambda I)I$. $\endgroup$ – user1551 Mar 2 at 13:20

I hope it's not too much trouble for me to talk in terms of endomorphisms rather than square matrices.

If we want to evaluate the polynomial $\text{det}(\phi - t I)$ at a matrix, it is natural to view it as an element of $\text{End}_{\mathbb{F}} (V) [t]$. The factorization $\text{det}(t I - \phi ) = \text{adj}(tI - \phi ) (tI - \phi)$ makes sense if we allow ourselves to think of $\text{adj}(tI - \phi)$ as a polynomial with endomorphisms as entries.

After we think of these as elements in $\text{End}_{\mathbb{F}} (V)[t]$, the product $\text{adj}(A - \lambda I) ( A - \lambda I)$ works just like any polynomial ring over a noncommutative ring:

$$\sum_{i = 1}^n \phi_i t^i \sum_{j = 1}^m \psi_j t^j = \sum_{i = 1}^{n + m} \sum_{j + k = i, 1 \leq j \leq n, 1 \leq k \leq m} \phi_j \circ \psi_k t^{i}$$

In fact, this language is convenient for the proof: we have an isomorphism $$\text{End}_{\mathbb{F}} (V) [t] \cong \text{End}_{\mathbb{F}} (V \otimes_{\mathbb{F}} \mathbb{F}[t])$$ In $\text{End}_{\mathbb{F}} (V \otimes_{\mathbb{F}} \mathbb{F}[t])$, we have a factorization $$ \text{det}(\mu_t - \phi) 1_V = \text{adj}(\mu_t - \phi) (\mu_t - \phi) $$ Where $\mu_t$ is multiplication by $t$. Under the isomorphism, this becomes a factorization $$ p(t) = f(t)(t - \phi) $$ where $p(t) \in \text{End}_{\mathbb{F}} (V)[t]$ is the characteristic polynomial.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.